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ok this helps, but when you say "suppose u is chosen randomly",this is where the problem arises. Isn't there a way to ask the riddle so that the notion of probability of victory makes sense, without encountering this problem (for instance agains an opponent who chooses the sequence). – Denis CommentedDec 12, 2013 at 13:56

But isn't this going to depend on how the opponent chooses the sequence? As Joel Hamkins notes in another comment, if the opponent always chooses the same sequence, then there is a strategy that gets the right answer each time. :-) – Alexander Pruss CommentedDec 12, 2013 at 14:33

yes but the point is that we can win again any strategy of the opponent, even if he chooses the sequence after we chose our (probabilistic) strategy. This way we avoid talking about probabilities on sequences. – Denis CommentedDec 12, 2013 at 15:09

In the probabilistic variant, I don't see that you can win against any strategy of the opponent. If we are making no probabilistic assumptions whatsoever, then in particular we are not assuming that our choice of index i is independent of the opponent's choice of numbers. – Alexander Pruss CommentedDec 17, 2013 at 14:47

Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice. – Denis CommentedDec 17, 2013 at 15:21

I was assuming that "independently" has the meaning it does in probability theory (P(AB)=P(A)P(B) and generalizations for σ-fields). But that does require a probabilistic description of the opponent's choice. Of course, one could mean "independently" here in some non-mathematical causal sense. (And there may be philosophical reason for doing this: fitelson.org/doi.pdf ) Still, mixing the probabilistic with nonprobabilistic concepts might lead to some difficulties, though. – Alexander Pruss CommentedDec 18, 2013 at 15:21